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Козя
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29 дек 2007 |
Пожалуйста, проверьте интеграл, т.к. ответ не совпадает с маткадовским:
![$ \int_{}^{}{\frac{tg x}{3cos^2 x +5 sin^2 x +4}dx}= \int_{}^{}{\frac{tg x}{3cos^2 x +5 (1-cos^2 x) +4}dx}= \int_{}^{}{\frac{tg x}{9-2cos^2 x}dx}= \int_{}^{}{\frac{tg x}{9-\frac{2}{1+tg^2 x}}dx}= \int_{}^{}{\frac{tg x(1+tg^2 x)}{7-9tg^2 x}dx}= [t=tg x; x=arctg t; dx=\frac{dt}{t^2+1}]= \int_{}^{}{\frac{t(1+t^2)}{(7-9t^2)} \frac{dt}{(1+t^2)}} = \int_{}^{}{\frac{tdt}{(7-9t^2)} } = - \frac{1}{2}* \frac{1}{9} \int_{}^{}{\frac{d(7-9t^2)}{(7-9t^2)} } = - \frac{1}{18} ln(7-9t^2) +C = - \frac{1}{18} ln(7-9tg^2 x) +C $](http://teacode.com/service/latex/latex.png?latex=+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B3cos%5E2+x+%2B5+sin%5E2+x+%2B4%7Ddx%7D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B3cos%5E2+x+%2B5+%281-cos%5E2+x%29+%2B4%7Ddx%7D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B9-2cos%5E2+x%7Ddx%7D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B9-%5Cfrac%7B2%7D%7B1%2Btg%5E2+x%7D%7Ddx%7D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%281%2Btg%5E2+x%29%7D%7B7-9tg%5E2+x%7Ddx%7D%3D+%5Bt%3Dtg+x%3B++x%3Darctg+t%3B++dx%3D%5Cfrac%7Bdt%7D%7Bt%5E2%2B1%7D%5D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Bt%281%2Bt%5E2%29%7D%7B%287-9t%5E2%29%7D+%5Cfrac%7Bdt%7D%7B%281%2Bt%5E2%29%7D%7D+%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btdt%7D%7B%287-9t%5E2%29%7D+%7D+%3D+-+%5Cfrac%7B1%7D%7B2%7D*+%5Cfrac%7B1%7D%7B9%7D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Bd%287-9t%5E2%29%7D%7B%287-9t%5E2%29%7D+%7D+%3D+-+%5Cfrac%7B1%7D%7B18%7D+ln%287-9t%5E2%29++%2BC+%3D+-+%5Cfrac%7B1%7D%7B18%7D+ln%287-9tg%5E2+x%29++%2BC+&fontsize=21 "$ \int_{}^{}{\frac{tg x}{3cos^2 x +5 sin^2 x +4}dx}= \int_{}^{}{\frac{tg x}{3cos^2 x +5 (1-cos^2 x) +4}dx}= \int_{}^{}{\frac{tg x}{9-2cos^2 x}dx}= \int_{}^{}{\frac{tg x}{9-\frac{2}{1+tg^2 x}}dx}= \int_{}^{}{\frac{tg x(1+tg^2 x)}{7-9tg^2 x}dx}= [t=tg x; x=arctg t; dx=\frac{dt}{t^2+1}]= \int_{}^{}{\frac{t(1+t^2)}{(7-9t^2)} \frac{dt}{(1+t^2)}} = \int_{}^{}{\frac{tdt}{(7-9t^2)} } = - \frac{1}{2}* \frac{1}{9} \int_{}^{}{\frac{d(7-9t^2)}{(7-9t^2)} } = - \frac{1}{18} ln(7-9t^2) +C = - \frac{1}{18} ln(7-9tg^2 x) +C $")
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Козя
#
29 дек 2007 |
С первого раза решение не отобразилось полностью, вот оно:
![$ \int_{}^{}{\frac{tg x}{3cos^2 x +5 sin^2 x +4}dx}= \int_{}^{}{\frac{tg x}{3cos^2 x +5 (1-cos^2 x) +4}dx} = \int_{}^{}{\frac{tg x}{9-2cos^2 x}dx} = \int_{}^{}{\frac{tg x}{9-\frac{2}{1+tg^2 x}}dx} = \int_{}^{}{\frac{tg x(1+tg^2 x)}{7-9tg^2 x}dx} = [t=tg x; x=arctg t; dx=\frac{dt}{t^2+1}]= \int_{}^{}{\frac{t(1+t^2)}{(7-9t^2)} \frac{dt}{(1+t^2)}} = \int_{}^{}{\frac{tdt}{(7-9t^2)} }$](http://teacode.com/service/latex/latex.png?latex=+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B3cos%5E2+x+%2B5+sin%5E2+x+%2B4%7Ddx%7D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B3cos%5E2+x+%2B5+%281-cos%5E2+x%29+%2B4%7Ddx%7D+%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B9-2cos%5E2+x%7Ddx%7D+%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%7D%7B9-%5Cfrac%7B2%7D%7B1%2Btg%5E2+x%7D%7Ddx%7D+%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btg+x%281%2Btg%5E2+x%29%7D%7B7-9tg%5E2+x%7Ddx%7D+%3D+%5Bt%3Dtg+x%3B++x%3Darctg+t%3B++dx%3D%5Cfrac%7Bdt%7D%7Bt%5E2%2B1%7D%5D%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Bt%281%2Bt%5E2%29%7D%7B%287-9t%5E2%29%7D+%5Cfrac%7Bdt%7D%7B%281%2Bt%5E2%29%7D%7D+%3D+%5Cint_%7B%7D%5E%7B%7D%7B%5Cfrac%7Btdt%7D%7B%287-9t%5E2%29%7D+%7D&fontsize=21 "$ \int_{}^{}{\frac{tg x}{3cos^2 x +5 sin^2 x +4}dx}= \int_{}^{}{\frac{tg x}{3cos^2 x +5 (1-cos^2 x) +4}dx} = \int_{}^{}{\frac{tg x}{9-2cos^2 x}dx} = \int_{}^{}{\frac{tg x}{9-\frac{2}{1+tg^2 x}}dx} = \int_{}^{}{\frac{tg x(1+tg^2 x)}{7-9tg^2 x}dx} = [t=tg x; x=arctg t; dx=\frac{dt}{t^2+1}]= \int_{}^{}{\frac{t(1+t^2)}{(7-9t^2)} \frac{dt}{(1+t^2)}} = \int_{}^{}{\frac{tdt}{(7-9t^2)} }$")
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О.А.
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29 дек 2007 |
маленькая арифметическая ошибка, после преобразований получим
 поэтому ответ
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